- #1

cmkluza

- 118

- 1

## Homework Statement

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 58.0 m away, making a 3.00° angle with the ground. How fast was the arrow shot?

## Homework Equations

Kinematic equations are useful, I think.

##d = v_it + \frac{1}{2}at^2##

## The Attempt at a Solution

As I wrote this out I realized a huge flaw in my logic - I assumed there to be no initial vertical velocity because the arrow was shot parallel, but then assumed the path the arrow took was a triangle right from the start (which would imply it's being shot at an angle, not parallel to the ground). Thus, my work below is definitely incorrect, but it's all I've got so far so I'll leave it. Where can I start with the given information?

I started by forming a triangle (angles 90, 87, and 3 degrees). Using the sine rule, I found the vertical height the arrow was shot from:

##\frac{h}{\sin(3)} = \frac{58}{\sin(87)} \longrightarrow h = \frac{58\sin(3)}{\sin(87)} \approx 3.04##m

It stated the arrow was shot parallel to the ground, so I assume an initial vertical velocity of zero. I then use acceleration of gravity (9.81) to try to solve for total time the arrow is in the air:

##d = v_it + \frac{1}{2}at^2 \longrightarrow 3.04 = \frac{1}{2}(9.81)t^2 \longrightarrow t^2 = \frac{2 \times 3.04}{9.81} \approx 0.62 \longrightarrow t = \sqrt{0.62} \approx 0.79##

With this, I assume I can solve for initial horizontal velocity, which (as it was shot parallel to the ground) should be the entire initial velocity, and there should be no horizontal acceleration:

##d = v_it + \frac{1}{2}at^2 \longrightarrow 58 = v_i(0.79) \longrightarrow v_i = \frac{58}{0.79} \approx 73.68##.

However, this is not the answer. Any hints on what I'm doing wrong?